Let $h(x)=x\cos^3(x)$. Find $h'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\cos^3(x)-x\sin^3(x)$ (Choice B) B $-3\cos^2(x)\sin(x)$ (Choice C) C $\cos^2(x)\left(\cos(x)-3x\sin(x)\right)$ (Choice D) D $\cos^2(x)\left(\cos(x)+3x\right)$
$h$ is a product of a function and a composite function. Let... $u(x)=x$ $v(x)=x^3$ $w(x)=\cos(x)$... then $h(x)=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $h'(x)$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=1$ $v'(x)=3x^2$ $w'(x)=-\sin(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=1\cdot{\cos^3(x)}+x\cdot{3\cos^2(x)}\cdot(-\sin(x)) \\\\ &=\cos^3(x)-3x\cos^2(x)\sin(x) \\\\ &=\cos^2(x)(\cos(x)-3x\sin(x)) \end{aligned}$ In conclusion, $h'(x)=\cos^2(x)\left(\cos(x)-3x\sin(x)\right)$.